<u who="nf0831"> on <pause dur="0.3"/> use of language in <pause dur="0.5"/> lectures </u><u who="om0832" trans="latching"> yes </u><u who="nf0831" trans="latching"> # so # </u><u who="om0832" trans="overlap"> for the benefit of international students </u><u who="nf0831" trans="latching"> sorry </u><pause dur="0.2"/> <u who="om0832" trans="pause"> for the benefit of international students </u><u who="nf0831" trans="overlap"> right <pause dur="0.4"/> # so there's no need to sort of adjust your hair or or worry that you wore the wrong clothes today <pause dur="1.2"/> # <pause dur="0.9"/> okay what i want to do today is <pause dur="1.2"/> to <pause dur="0.5"/> # just go over again this idea of a a thermal resistance model <pause dur="0.7"/> and why it's so useful because <pause dur="0.3"/> i think your group like <pause dur="0.2"/> # <pause dur="1.3"/> # many <pause dur="0.5"/> sort of groups at this stage think this is getting all far too complicated and there are too many equations and i don't know what goes where <pause dur="1.2"/><kinesic desc="puts on transparency" iterated="n"/> # <pause dur="1.2"/> i'll go back to <pause dur="0.2"/> # an example <pause dur="0.2"/> we looked at <pause dur="0.9"/> # earlier in the course of just what a real <pause dur="0.2"/> heat transfer problem might <pause dur="0.5"/> get to <pause dur="0.8"/> this isn't <pause dur="0.2"/> a particularly complicated system <pause dur="0.5"/> it's just <pause dur="0.4"/> # a sort of a building wall with a window in it so something that you <pause dur="0.3"/> you come across every day <pause dur="0.9"/> but if you look at it in heat transfer terms <pause dur="0.4"/> there's many different sort of pathways that the heat can go <pause dur="0.8"/> so you've got a a sort of a channel through here through the brick through the cavity insulation brick <pause dur="0.4"/> with <trunc>conduc</trunc> convection and radiation <pause dur="0.4"/> to the surface from the inside <pause dur="0.3"/> from the surface from the outside <pause dur="0.5"/> you've got a direct brick path here along the window frame <pause dur="0.4"/> you've got a conduction path through the

actual # <trunc>so</trunc> sorry on the on the window surround <pause dur="0.4"/> you've got a a <pause dur="0.3"/> path through the window frame <pause dur="0.8"/> there's the path through the window the glass <pause dur="0.3"/> whatever is in the gap <pause dur="0.4"/> the glass on the other side <pause dur="0.4"/> and all the same down here <pause dur="0.6"/> so trying to analyse this in heat transfer terms <pause dur="0.4"/> is rather difficult <pause dur="1.2"/> i've drawn a resistance model out really just to the <trunc>t</trunc> of of <pause dur="0.2"/> # <pause dur="0.4"/> essentially the top <kinesic desc="changes transparency" iterated="y" dur="4"/> half of this because it's symmetrical <pause dur="0.5"/> to show all the different sort of thermal resistances that you might put on this <pause dur="0.6"/> so here is the convection radiation <pause dur="0.2"/> from the inside surface <pause dur="0.4"/> here is the convection and radiation to the outside surface <pause dur="0.5"/> and here are the different heat transfer paths in thermal resistance <pause dur="0.2"/> notation so brick insulation brick <pause dur="0.6"/> direct brick <pause dur="0.2"/> directly through the frame <pause dur="0.5"/> glass convection and radiation these will be different coefficients between the two plate panes of glass <pause dur="0.4"/> and glass <pause dur="0.2"/> on the other side <pause dur="0.9"/> okay <pause dur="0.2"/> the reason we go for that <pause dur="0.3"/> # <pause dur="0.2"/> approach <pause dur="0.4"/> rather than trying to

actually calculate individual <pause dur="0.4"/> # components so the reason why we put everything into a thermal <pause dur="0.3"/> resistance form <pause dur="0.7"/> is that once you've done that <pause dur="0.8"/> you've basically got a very simple equation if that was electrical resistances you could add them up <pause dur="0.3"/> you wouldn't need very complicated <pause dur="0.3"/> # methods to do it it isn't like <pause dur="0.4"/> # some of the # matrix circuits that you've looked at that's just a straightforward electrical resistance circuit <pause dur="0.6"/> so you can add that up <pause dur="0.7"/> you can write it as a total thermal resistance <pause dur="3.3"/> like that <pause dur="1.7"/><vocal desc="cough" iterated="n"/><pause dur="1.2"/> with the <pause dur="0.2"/> inside here <pause dur="2.3"/> and the outside here <pause dur="2.9"/> so this is the total thermal resistance <pause dur="4.0"/> and that's the <pause dur="0.2"/> driving force like voltage in an electrical circuit <pause dur="0.6"/> so the reet <trunc>h</trunc> rate of head transfer Q-dot <pause dur="1.1"/> is just the temperature difference T-in-minus-T-out <pause dur="2.9"/> over R-total <pause dur="4.6"/> and when you write out <pause dur="0.2"/> a heat transfer problem in that sort of form <pause dur="1.3"/> what it enables you to do is to see which <pause dur="0.4"/> # components in that <pause dur="0.4"/> are going to be important <pause dur="0.3"/> for the overall <pause dur="0.2"/> heat transfer <pause dur="0.8"/> so you may have

to use you almost certainly will have to use <pause dur="0.2"/> approximations <pause dur="0.3"/> to find <pause dur="0.2"/> these thermal resistances <pause dur="0.2"/> the conduction ones are quite straightforward convection and radiation as we've looked at are more complicated <pause dur="0.7"/> but once you've got them and you've got some numbers on the diagram <pause dur="0.4"/> you can decide which are the sort of dominant <pause dur="0.2"/> # issues <pause dur="0.7"/> and <pause dur="0.2"/> if <pause dur="0.2"/> something isn't going to be important in the overall heat transfer <pause dur="0.4"/> for instance in this problem if you look at the thermal resistance of the glass <pause dur="0.4"/> it's very very small so in terms of changing the heat transfer <pause dur="0.3"/> it really wouldn't matter if you made the glass a bit thicker or if you changed to a different sort of <pause dur="0.3"/> glass <pause dur="0.9"/> on the other hand when you get to modern double glazed windows where <pause dur="0.3"/> the actual glass part of it <pause dur="0.3"/> has been # optimized to reduce heat loss <pause dur="0.5"/> you may suddenly find that the dominant channel of heat transfer is going to be through the <pause dur="0.2"/> metal frame <pause dur="2.6"/> so the reason for doing this is to get <pause dur="0.2"/> what may not be a completely correct

starting point but at least an approximate starting point <pause dur="0.4"/> where you can <pause dur="0.2"/> approach <pause dur="0.2"/> a real problem <pause dur="3.4"/> the <pause dur="1.2"/><kinesic desc="changes transparency" iterated="y" dur="15"/> driving mechanisms that we looked at the mechanisms that you've got to get into <pause dur="0.4"/> heat transfer format <pause dur="0.4"/> are conduction that we've looked at <pause dur="0.3"/> # quite a bit earlier on either for flat plates or for cylinders <pause dur="0.5"/> and both of them you can just add them up in series <pause dur="0.4"/> # you can add them up in parallel as well although we haven't looked specifically at that <pause dur="1.0"/> convection <pause dur="0.5"/> where the convected coefficient which determines the <pause dur="0.3"/> # <pause dur="0.2"/> convective <pause dur="0.3"/> resistance <pause dur="0.4"/> # you need <pause dur="0.2"/> to go through the procedures we've looked at over the last two weeks <pause dur="0.4"/> either free convection or forced convection <pause dur="0.4"/> # and working out <pause dur="0.2"/> how this relates to the fundamental properties <pause dur="0.3"/> of the fluid <pause dur="0.9"/> and the difficult one and the one that doesn't really fit <pause dur="0.2"/> into this linear model <pause dur="0.3"/> radiation <pause dur="1.2"/> and the reason why we ploughed through last week looking at how we could write radiation into an approximate linear form <pause dur="0.5"/> is so that radiation can

fit into a resistance model <pause dur="0.3"/> like everything else <pause dur="0.3"/> so that instead of having to write radiation as a fourth power relationship <pause dur="0.4"/> we can write it <pause dur="0.2"/> at least approximately <pause dur="0.3"/> as a linear relationship <pause dur="0.4"/> then <pause dur="0.3"/> radiation is described by a thermal resistance <pause dur="0.2"/> and a temperature difference <pause dur="0.3"/> to a first approximation <pause dur="0.3"/> and then you can just slip it in <pause dur="0.3"/> in the model like everything else <pause dur="0.5"/> and it's exactly the same in an electrical circuit if you had a component <pause dur="0.3"/> which didn't have a linear relationship between <pause dur="0.3"/> current and voltage <pause dur="0.4"/> or i mean you can in heat transfer like in electricity you can have # # things that are out of phase you can use complex numbers to represent thermal storage in elements but <pause dur="0.3"/> fortunately i'm not going to do that <pause dur="0.5"/> if you have something that doesn't have a linear relationship between <trunc>co</trunc> # the the current and voltage <pause dur="0.3"/> then it's much more difficult to incorporate it into a <trunc>sing</trunc> a simple <pause dur="0.2"/> linear model <pause dur="0.6"/> yes </u><u who="sm0833" trans="overlap"> can i have another look at that model <gap reason="inaudible" extent="2 secs"/> </u><pause dur="1.0"/> <u who="nf0831" trans="pause"> the <pause dur="0.2"/> <trunc>th</trunc> <pause dur="0.3"/> this one <kinesic desc="changes transparency" iterated="y" dur="3"/></u><pause dur="0.7"/> <u who="sm0833" trans="pause"> yeah </u><pause dur="1.7"/> <u who="nf0831" trans="pause"> it was meant

as a sketch rather than a a sort of something to <pause dur="0.2"/> take down in detail really just to show how you can get <pause dur="0.3"/> all the bits together </u><pause dur="3.0"/> <u who="sm0834" trans="pause"> can you put that sheet back <gap reason="inaudible" extent="1 sec"/><vocal desc="laugh" iterated="n"/></u><pause dur="0.2"/> <u who="nf0831" trans="pause"> pardon </u><pause dur="0.2"/> <u who="sm0834" trans="pause"> <gap reason="inaudible" extent="1 sec"/> sheet back up <gap reason="inaudible" extent="1 sec"/></u><pause dur="2.8"/> <u who="nf0831" trans="pause"> i can't get them both on at the same time i <trunc>onl</trunc> i only wanted </u><u who="sm0835" trans="overlap"> <gap reason="inaudible" extent="2 secs"/></u><pause dur="0.6"/> <u who="nf0831" trans="pause"> # <pause dur="0.6"/> that one we have looked at several times before <pause dur="0.3"/> # i can't get them both on at the same time <pause dur="0.4"/> # so we we will have a quick sketch a a model sketch and # i'll leave that up whilst i say the next # parts of what i want to do <pause dur="0.9"/> so really what i'm saying is <pause dur="0.2"/> you can make <pause dur="0.4"/> in principle quite a complicated model <pause dur="0.5"/> but provided everything is linear <pause dur="0.3"/> it just simplifies to <pause dur="0.2"/> total resistance <pause dur="0.3"/> heat flux is total is the <trunc>temp</trunc> overall temperature difference <pause dur="0.2"/> divided by the total <pause dur="0.2"/> resistance <pause dur="0.4"/> and then when you've <trunc>est</trunc> got you've worked through to that level <pause dur="0.3"/> you then start worrying about which of these components are important and where you need to know things <pause dur="0.2"/> more accurately <pause dur="0.8"/> and once you get to real problems there's obviously some heat transfer <pause dur="0.4"/> # going

perpendicular to the direction of the temperature differences odd edge effects around corners <pause dur="0.4"/> but at least if you've got a first stab you know <pause dur="0.2"/> which bits are important to look at <pause dur="0.3"/> and which aren't <pause dur="2.2"/> okay today i want to look at two things <pause dur="0.4"/> # the first is just to look at the second problem <pause dur="0.2"/> from the <pause dur="0.3"/> problem sheet from the tutorial last week which i didn't go over in the tutorial <pause dur="0.6"/> and then to start building up this rather open-ended problem <pause dur="0.4"/> of designing the # cooling system for <pause dur="0.3"/> # the departmental <pause dur="0.3"/> store <pause dur="0.2"/> in time for the <pause dur="0.2"/> strawberries at Wimbledon <pause dur="1.0"/> so we're going # if you want to be there's have you finished <pause dur="0.4"/> <trunc>m</trunc> <pause dur="0.2"/> resistance model sketching <pause dur="0.4"/> i'll just leave that up <pause dur="0.5"/> for the moment <pause dur="0.7"/> # so problem two in the heat transfer # sheet <pause dur="0.2"/> number four <pause dur="1.9"/> oh i'm just raise that slightly <event desc="adjusts equipment" iterated="y" dur="10"/> so i've got a bit of board <pause dur="1.0"/> at the <pause dur="0.2"/> bottom <pause dur="4.3"/> reason i'd hoped this would be on video is to to show people how difficult it is if you don't have a separate whiteboard and a <pause dur="0.2"/> # an overhead projector but # <pause dur="0.2"/> that will have to be for the

next one <pause dur="4.4"/> okay so the second the the second problem <pause dur="0.5"/> is one where in fact you don't have to it's a it's a very straightforward problem you don't have to do any <pause dur="0.3"/> approximations <pause dur="0.6"/> you're designing <pause dur="0.2"/> an insulation system for a furnace wall i'll do this down here and then move it up in a minute <pause dur="0.4"/> so this is # sheet four number two <pause dur="7.8"/><kinesic desc="writes on board" iterated="y" dur="6"/>

you've got a furnace wall at a thousand degrees C <pause dur="6.3"/><kinesic desc="writes on board" iterated="y" dur="3"/> and you <pause dur="0.3"/> # have got <pause dur="0.3"/> # <pause dur="0.7"/> an <pause dur="0.2"/> environment around at forty <pause dur="0.2"/><kinesic desc="writes on board" iterated="y" dur="2"/> degrees C <pause dur="2.8"/> a nasty hot corner of a factory <pause dur="3.8"/> you've been asked to insulate <pause dur="0.2"/> the furnace wall and you've got two <pause dur="0.4"/> different types of <pause dur="0.2"/> insulation <pause dur="1.3"/> i'm not showing these <pause dur="0.5"/> necessarily to the right relative thickness <pause dur="0.9"/> you're starting with mineral wall because that will withstand high temperatures <pause dur="0.4"/><kinesic desc="writes on board" iterated="y" dur="19"/> and this has got a thermal conductivity of seventy <pause dur="1.8"/> milliwatts <pause dur="3.1"/> per metre <pause dur="2.1"/> per degree kelvin <pause dur="1.8"/> and you're following it with fibreglass <pause dur="2.9"/> which has got a better thermal conductivity so it's <trunc>ge</trunc> better insulation material <pause dur="0.5"/> but it can't go up to such high temperatures <pause dur="0.4"/> so it's to # it's got a thermal conductivity of forty <pause dur="0.3"/> milliwatts per metre <kinesic desc="writes on board" iterated="y" dur="7"/> per degree kelvin <pause dur="10.8"/> and <pause dur="0.5"/> conveniently somebody has estimated for you the heat transfer coefficient <pause dur="0.4"/> from the surface <pause dur="0.4"/> so this sort of first estimate of you know roughly what temperature it is so what heat transfer coefficient

is it going to be <pause dur="0.4"/> has been done for you <pause dur="0.6"/><kinesic desc="writes on board" iterated="y" dur="3"/> so H <pause dur="0.2"/> for the surface <pause dur="0.5"/> has been estimated <pause dur="0.3"/> at about fifteen watts per square metre <pause dur="0.2"/><kinesic desc="writes on board" iterated="y" dur="5"/> per degree kelvin <pause dur="4.6"/> and that includes both convection and radiation it's the overall heat loss from the surface <pause dur="14.3"/><kinesic desc="writes on board" iterated="y" dur="9"/> and you've got two <pause dur="0.4"/> # <pause dur="0.4"/> criteria two things to satisfy <pause dur="0.3"/> on this <pause dur="0.6"/> the first thing that you must satisfy is that the outer surface <pause dur="0.3"/> mustn't be more than <pause dur="0.2"/> fifty-five degrees celsius <pause dur="0.3"/> can i take the # <pause dur="0.2"/> overhead off now </u><pause dur="0.3"/> <u who="sm0836" trans="pause"> <gap reason="inaudible" extent="1 sec"/></u><u who="nf0831" trans="latching"> 'cause it means i don't have to go around on my knees <pause dur="8.4"/><event desc="puts away screen" iterated="n"/><event desc="turns off overhead projector" iterated="n"/> so the conditions that this system has got to filfil fulfil <pause dur="0.5"/> is that the temperature at the interface between <pause dur="0.2"/> the two media <pause dur="0.4"/> has got to be no more than four-hundred <pause dur="0.2"/> degrees celsius <pause dur="8.6"/><kinesic desc="writes on board" iterated="y" dur="8"/> and the temperature at the outside surface <pause dur="0.3"/> has to be not more than fifty-five <pause dur="0.2"/><kinesic desc="writes on board" iterated="y" dur="10"/> degrees celsius <pause dur="8.8"/> so the surface temperature limitation is so that people don't hurt their hands when they touch it <pause dur="0.4"/> or bump into it by mistake <pause dur="0.4"/> rather high for a surface environment that people are going to be close but it's

obviously a very hot <pause dur="0.2"/> environment that it's in <pause dur="0.6"/> this interface temperature <pause dur="0.4"/> the reason there is so that the outer insulation material doesn't work at a temperature <pause dur="0.3"/> that it doesn't like to be at <pause dur="0.8"/> and so what you're asked to do is to calculate the thicknesses <pause dur="0.4"/> of the <pause dur="0.2"/> two insulation materials or the minimum thicknesses <pause dur="0.2"/> to achieve these <pause dur="0.2"/> obviously the more insulation you put on <pause dur="0.3"/> the lower these two temperatures will go the more <pause dur="0.2"/> these surfaces will approach <pause dur="0.3"/> the temperature <pause dur="0.3"/> of the surroundings <pause dur="11.0"/><kinesic desc="writes on board" iterated="y" dur="10"/> okay <pause dur="0.2"/> how do you think you can go about <pause dur="0.4"/> doing that <pause dur="1.5"/> any suggestions <pause dur="0.6"/> this is an occasion where you don't have to make any approximations any assumptions <pause dur="0.4"/> you've got all the information there to do an exact calculation <pause dur="0.5"/> if you take this <pause dur="0.3"/> # assumed value for the surface <pause dur="0.2"/> heat transfer coefficient <pause dur="4.9"/> yes </u><u who="sf0837" trans="overlap"> <gap reason="inaudible" extent="4 secs"/> </u><pause dur="0.3"/> <u who="nf0831" trans="pause"> have i indeed <pause dur="0.8"/> thank you <pause dur="1.3"/> that's a very good starting point is to actually get your data correct <pause dur="6.6"/><kinesic desc="writes on board" iterated="y" dur="2"/> in fact fifteen is a is a more sort of realistic value that you're likely to get in a <pause dur="0.4"/> #

with convection and radiation combined thank you <pause dur="2.2"/> okay what do you need to know <pause dur="1.2"/> in order to calculate <pause dur="0.3"/> one or both of these <pause dur="2.8"/> so the thicknesses here i can call it X-A <pause dur="8.7"/><kinesic desc="writes on board" iterated="y" dur="9"/> and X-B <pause dur="6.1"/> you could <pause dur="0.4"/> for instance write down the equation including one of those and then decide what you knew in the equation <pause dur="0.3"/> and what you didn't know </u><pause dur="3.2"/> <u who="sm0838" trans="pause"> Q-N through the material </u><pause dur="0.7"/> <u who="nf0831" trans="pause"> sorry </u><u who="sm0838" trans="overlap"> 'cause <pause dur="0.4"/> the conduction through the material </u><pause dur="0.3"/> <u who="nf0831" trans="pause"> yeah </u><pause dur="0.6"/> <u who="sm0838" trans="pause"> is going to be <pause dur="0.8"/> the radiation and convection </u><pause dur="0.6"/> <u who="nf0831" trans="pause"> that's right so you know that the conduction through the wall <pause dur="12.8"/><kinesic desc="writes on board" iterated="y" dur="14"/> is equal to the <pause dur="1.3"/> i'll do it with a Q-dot to say that it's a rate <pause dur="0.9"/><kinesic desc="writes on board" iterated="y" dur="1"/> a <pause dur="0.5"/> flow per unit time is equal to the heat flux <pause dur="0.3"/> # Q-dot <kinesic desc="writes on board" iterated="y" dur="18"/> loss <pause dur="3.5"/> from the surface <pause dur="13.4"/> so you're making the usual assumption that you're in steady state <pause dur="0.8"/> so that's a good starting point <pause dur="7.6"/> having made that start <pause dur="0.2"/> what can you do next <pause dur="19.2"/> there's no <pause dur="0.3"/> tricks there's no <pause dur="0.3"/> # # there's there's no <pause dur="0.2"/> assumptions in this you've just got to decide <pause dur="0.4"/> how you can find a value for this thickness <pause dur="0.6"/> and this thickness and once you've done one it'll be perfectly obvious how to do the <pause dur="0.4"/> the other one </u><pause dur="5.1"/> <u who="sm0839" trans="pause"> do you need to get a value of the <pause dur="0.9"/> each resistance </u><pause dur="1.2"/> <u who="nf0831" trans="pause"> you need to get a

value of each resistance so you could say that Q-dot-N <pause dur="4.0"/><kinesic desc="writes on board" iterated="y" dur="5"/> is the <pause dur="0.2"/> temperature difference <pause dur="0.6"/> # we've used A and B for the materials so if i call this sort of <kinesic desc="writes on board" iterated="y" dur="6"/> T-one <pause dur="1.0"/> T-two <pause dur="2.2"/> T-three <pause dur="2.1"/> we can say that Q-dot <pause dur="0.2"/><kinesic desc="writes on board" iterated="y" dur="16"/> is equal to T-one minus T-two <pause dur="1.4"/> over <pause dur="0.8"/> R <pause dur="0.2"/> the thermal resistance of layer A <pause dur="1.9"/> which is equal to <pause dur="0.6"/> T-two minus T-three <pause dur="0.8"/> over the thermal resistance of layer B <pause dur="1.8"/> how would you calculate the thermal resistance <pause dur="0.4"/> of each layer </u><pause dur="10.5"/> <u who="sm0840" trans="pause"> i think it's something <pause dur="0.5"/> multiplied by the <pause dur="1.4"/> K-A </u><pause dur="1.3"/> <u who="nf0831" trans="pause"> # <pause dur="0.4"/> the thermal <kinesic desc="writes on board" iterated="y" dur="3"/> resistance is the thickness <pause dur="1.5"/> any advance on multiply </u><pause dur="3.6"/> <u who="sm0841" trans="pause"> divide </u><pause dur="0.5"/> <u who="nf0831" trans="pause"> divide <pause dur="3.4"/><kinesic desc="writes on board" iterated="y" dur="7"/><vocal desc="laughter" iterated="y" dur="1"/> okay the thickness is X divided by <pause dur="0.3"/> K <pause dur="0.7"/> and one more </u><pause dur="0.6"/> <u who="sm0842" trans="pause"> A </u><pause dur="0.4"/> <u who="nf0831" trans="pause"> A that's right so <pause dur="0.2"/> the thickness is X over K-A <pause dur="4.6"/><kinesic desc="writes on board" iterated="y" dur="4"/> so are you getting any <pause dur="0.3"/> closer <pause dur="1.1"/> to <pause dur="0.5"/> # <pause dur="0.3"/> a solution on this so you know that Q-dot-N <pause dur="0.3"/> is equal to Q-dot-loss <pause dur="0.6"/> you know that the same Q-dot-N is going through each <pause dur="0.2"/> layer <pause dur="0.6"/> so that you can write to the conduction equation for the first layer <pause dur="0.3"/> and for the second layer <pause dur="0.3"/> and in each case for each layer the <pause dur="0.3"/> # the X-over-K-A is equal to the <pause dur="0.5"/> resistance <pause dur="2.3"/> and the things that you know <pause dur="0.5"/> on this

'cause they give it in the problem you know the temperatures <pause dur="0.8"/> 'cause that's what your <pause dur="0.3"/> criteria that you're trying to set <pause dur="0.8"/> you know <pause dur="1.2"/> the <pause dur="0.2"/> K value <pause dur="2.2"/> one problem with this is that nowhere in the <pause dur="0.3"/> question have you got anything <pause dur="0.4"/> to do with the area <pause dur="1.0"/> so you certainly don't know <pause dur="0.3"/> the area <pause dur="2.0"/> and there's no information given from which you can calculate <pause dur="0.4"/> the area <pause dur="1.3"/>

and if you think about it it shouldn't really matter <pause dur="0.4"/> 'cause you're <trunc>defi</trunc> you're designing something for a building wall <pause dur="0.3"/> and you know it's got to have certain temperatures <pause dur="0.3"/> it shouldn't matter what area it is because you've got to make sure it's got that temperatures whether you make it big or small so </u><pause dur="0.9"/> <u who="sm0843" trans="pause"> can could you work it out <pause dur="0.6"/> per unit area </u><u who="nf0831" trans="latching"> you can work it out per unit area so that's that's the way that i would approach this <pause dur="0.3"/><kinesic desc="writes on board" iterated="y" dur="1"/> you can say that you can't actually do the calculations with an area and you don't you can't put a number in for the area <pause dur="0.3"/> so you can work <pause dur="0.2"/> per <kinesic desc="writes on board" iterated="y" dur="4"/> unit area <pause dur="0.8"/> so we've established this we've established this <pause dur="0.4"/> we've established that you need to work per unit area <pause dur="8.1"/><kinesic desc="writes on board" iterated="y" dur="14"/> 'cause there's no value <pause dur="3.6"/> for A <pause dur="0.3"/> and it shouldn't matter <pause dur="0.4"/> what the value of A is you should have the same <pause dur="0.3"/> temperatures at the interface if you make the <trunc>ar</trunc> the <pause dur="0.2"/> the wall twice as big in area or half as big <pause dur="0.3"/> in area <pause dur="1.4"/> okay so you're still left with the problem though you know <pause dur="0.3"/> you're going to work per unit area <pause dur="0.4"/> you want to know the <pause dur="0.3"/> value <pause dur="0.4"/> of X <pause dur="0.6"/> you know the temperatures <pause dur="0.3"/> we've established you don't need the area you'll work per unit area <pause dur="0.8"/> how can you find

though the heat flux so the only thing you're really left with in one of these equations <pause dur="0.3"/> is to find the heat flux <pause dur="0.3"/> per unit area <pause dur="1.3"/> how can you do that <pause dur="14.4"/> and <pause dur="0.2"/> it looks like you could do it either for the heat loss from the surface or for the conduction through the wall <pause dur="0.5"/> because they're <pause dur="0.6"/> the same <pause dur="2.0"/> so <pause dur="0.3"/> can you do either of those which one can you do <pause dur="13.9"/> anyone going to hazard a <pause dur="0.2"/> a try <pause dur="5.1"/> how do you calculate <pause dur="0.8"/> heat loss from the surface you you you've established that to calculate the conduction <pause dur="0.5"/> you need to know <pause dur="0.7"/> the property that you've got to calculate <pause dur="0.5"/> so you can't calculate the conduction flux <trunc>direc</trunc> directly <pause dur="0.7"/> how do you calculate the heat loss <pause dur="0.3"/> from the surface </u><pause dur="11.8"/> <u who="sm0844" trans="pause"> <gap reason="inaudible" extent="1 sec"/> <pause dur="0.2"/> <vocal desc="clears throat" iterated="n"/> <gap reason="inaudible" extent="2 secs"/> </u><pause dur="0.8"/> <u who="nf0831" trans="pause"> yes </u><u who="sm0844" trans="latching"> <gap reason="inaudible" extent="1 sec"/></u><u who="nf0831" trans="overlap"> now you've got a <trunc>con</trunc> coefficient that's right so you've got a coefficient that <pause dur="0.5"/> tells you about both convection and radiation the sum of them <pause dur="1.0"/> so what equation can you do to write down <pause dur="0.8"/> to give you the rate of heat loss from the surface </u><u who="sm0845" trans="latching"> T-S minus T-A <pause dur="0.4"/> <gap reason="inaudible" extent="1 sec"/></u><pause dur="0.4"/> <u who="nf0831" trans="pause"> that's right so you can say that Q-<sic corr="dot">lot</sic>-loss <pause dur="3.2"/><kinesic desc="writes on board" iterated="y" dur="15"/> is the <pause dur="0.2"/> temperature

difference <pause dur="1.2"/> divided by <pause dur="0.2"/> the <pause dur="0.2"/> surface resistance <pause dur="3.2"/> where <pause dur="0.3"/> R-S <pause dur="3.0"/> you could either you can either write it as a surface resistance or the other way up as a surface heat transfer coefficient <pause dur="0.5"/> in this term <pause dur="0.3"/> because you've got the surface heat transfer coefficient <pause dur="0.3"/> it's easier to say that R-S is <kinesic desc="writes on board" iterated="y" dur="3"/> one-<pause dur="0.3"/>over-A <pause dur="0.7"/> multiplied by <pause dur="0.4"/> the <pause dur="0.5"/> heat transfer coefficient for the surface <pause dur="0.7"/> so <kinesic desc="writes on board" iterated="y" dur="10"/> Q-dot-loss <pause dur="0.2"/> is <pause dur="0.4"/> A <pause dur="0.8"/> multiplied by H-S <pause dur="1.6"/> # <pause dur="0.2"/> multiplied by <pause dur="0.6"/> T-S-<pause dur="0.6"/>minus-T-A <pause dur="5.2"/> so finally you're getting somewhere that you can actually get some <pause dur="0.3"/> # numbers out <pause dur="0.3"/> you can't get a value for A there's nothing given to give you a value A <pause dur="0.4"/> but if you calculate Q-dot-loss <pause dur="0.2"/> over A <pause dur="9.5"/><kinesic desc="writes on board" iterated="y" dur="6"/> so the heat loss per unit area <pause dur="2.3"/><kinesic desc="writes on board" iterated="y" dur="6"/> is therefore H-S multiplied by T-S-minus-T-A <pause dur="0.4"/> sorry H-S has just gone off the top of the board as it always does when i need it <pause dur="0.3"/> so that's fifteen <pause dur="0.2"/><kinesic desc="writes on board" iterated="y" dur="6"/> watts <pause dur="0.8"/> per square metre <pause dur="1.0"/> per degree kelvin <pause dur="1.4"/> multiplied by the temperature difference fifty-five minus forty <pause dur="5.1"/><kinesic desc="writes on board" iterated="y" dur="12"/> which is therefore fifteen multiplied by <pause dur="0.4"/> fifteen <pause dur="0.3"/> which is

two-hundred-and-twenty-five </u><pause dur="2.2"/> <u who="sm0846" trans="pause"> it's not fifty-five minus forty kelvin </u><pause dur="1.4"/> <u who="nf0831" trans="pause"> # </u><u who="sm0847" trans="overlap"> <gap reason="inaudible" extent="2 secs"/></u><u who="nf0831" trans="latching"> okay it's <trunc>i</trunc> it's fifty-five it's fifty-five of of if you if you get a temperature interval <pause dur="0.6"/> i should have said it fifty-five-minus-forty <pause dur="0.6"/> measured in units of degrees kelvin or in degrees celsius 'cause the temperature interval is the same <pause dur="0.3"/> whether you're in kelvin <pause dur="0.4"/> or celsius <pause dur="3.8"/> so <pause dur="1.0"/> when it's an absolute temperature you're you're completely right you've got to be certain whether you're in kelvin or celsius <pause dur="0.3"/> when it's a temperature difference because the interval is the same <pause dur="0.4"/> # <pause dur="0.2"/> the unit <pause dur="0.2"/> can be either <pause dur="1.0"/> so you've got fifteen watts per square metre per degree kelvin <pause dur="0.4"/> multiplied by a temperature interval <pause dur="0.3"/> of fifteen kelvin <pause dur="0.3"/> or two-hundred-and-twenty-five <kinesic desc="writes on board" iterated="y" dur="5"/> watts <pause dur="0.4"/> per <pause dur="0.2"/> square metre <pause dur="5.5"/> okay so you know the heat loss <pause dur="0.2"/> per unit area <pause dur="0.4"/> and if the heat <pause dur="0.4"/> loss is equal to the heat conduction then the heat loss per unit area <pause dur="0.3"/> must be equal to the heat conduction <pause dur="0.4"/> per <pause dur="0.2"/> unit area <pause dur="0.8"/> so we can therefore say <pause dur="4.8"/><kinesic desc="writes on board" iterated="y" dur="5"/> the heat conduction <pause dur="0.4"/> per unit area <pause dur="0.3"/> through either of <kinesic desc="writes on board" iterated="y" dur="8"/> of the media <pause dur="1.9"/>

is also two-hundred-and-twenty-five <pause dur="0.3"/> watts <pause dur="0.8"/> per square metre <pause dur="5.5"/> so finally we've said that the heat conduction <pause dur="0.5"/> is the temperature difference divided by <pause dur="0.2"/> the thermal resistance <pause dur="0.4"/> and we've said that the thermal resistance is X-<pause dur="0.3"/>over-<pause dur="0.2"/>K-A <pause dur="2.1"/> so the heat conduction per unit area <pause dur="2.3"/> is therefore for the first material <pause dur="0.4"/> equal to the temperature difference <pause dur="6.4"/><kinesic desc="writes on board" iterated="y" dur="14"/> divided by <pause dur="0.8"/> X <pause dur="1.7"/> and multiplied by <pause dur="0.3"/> K <pause dur="0.2"/> for that <pause dur="0.4"/> medium <pause dur="1.4"/> so we're dividing by <pause dur="0.5"/> X-over-K <pause dur="1.1"/> and we've taken the A on to this side because we're working with everything <pause dur="0.4"/> per unit area <pause dur="2.8"/> so finally we have an equation where the only unknown <pause dur="0.3"/> is a thing that you want to calculate the thickness <pause dur="0.3"/> of this <pause dur="0.2"/> material <pause dur="1.6"/> this is for medium one the first one so i'll write this as X-A <pause dur="0.4"/> and K-A <pause dur="1.3"/> so the thickness of the <pause dur="0.4"/> mineral wall <pause dur="12.7"/><kinesic desc="writes on board" iterated="y" dur="13"/> X-A just <pause dur="0.3"/> cross-multiplying that equation <pause dur="0.5"/><kinesic desc="writes on board" iterated="y" dur="3"/> is therefore equal to the temperature <pause dur="0.2"/> difference <pause dur="0.7"/> that's the temperature difference <pause dur="0.2"/> i'll just put this down and see if the diagram will reappear <pause dur="1.0"/><kinesic desc="reveals covered diagram" iterated="n"/> yeah there's a temperature difference across the mineral wall

so it's a thousand minus four-hundred <pause dur="7.6"/><kinesic desc="writes on board" iterated="y" dur="4"/> again it's an interval so we can write it as <kinesic desc="writes on board" iterated="y" dur="3"/> kelvin or degrees celsius <pause dur="3.7"/> multiplied by the <kinesic desc="writes on board" iterated="y" dur="15"/> thermal conductivity of that medium <pause dur="0.4"/> seventy milliwatts per metre per degree kelvin so <pause dur="0.4"/> point <trunc>ne</trunc> zero-seven <pause dur="1.7"/> watts per metre <pause dur="0.9"/> per degree kelvin <pause dur="3.0"/> and divided by <pause dur="0.4"/> the <pause dur="0.2"/> heat flux per unit area <pause dur="0.7"/><kinesic desc="writes on board" iterated="y" dur="6"/> two-hundred-and-twenty-<pause dur="0.3"/>five <pause dur="1.8"/> watts <pause dur="0.2"/> per square metre <pause dur="3.6"/> okay with a bit of luck that's going to end up with the right dimensions <pause dur="0.4"/> # although it takes a long time i always put dimensions in equations 'cause then you can check <pause dur="0.3"/> if you've lost a term <pause dur="0.4"/> so the <pause dur="0.4"/> K cancel out with the <pause dur="0.5"/> K-to-the-minus-one <pause dur="0.6"/> watts cancel out <pause dur="1.2"/> and you're left with metres-to-the-minus-one divided by metres-to-the-minus-two <pause dur="0.3"/> or therefore metres <pause dur="0.2"/> which is what you would want for a <pause dur="0.4"/> thickness <pause dur="0.4"/> so if you can just <pause dur="0.3"/> # do the calculation on that <pause dur="43.2"/> a number </u><pause dur="1.0"/> <u who="sm0847" trans="pause"> point-one-eight-seven </u><pause dur="0.3"/> <u who="nf0831" trans="pause"> sorry </u><pause dur="0.3"/> <u who="sm0847" trans="pause"> point-one-eight-seven </u><u who="nf0831" trans="latching"> point-one-eight-seven that's that's all right i <pause dur="0.2"/> put left behind where i did mine i calculated <trunc>point-one-n</trunc>-nine my number <pause dur="0.5"/> so that's <kinesic desc="writes on board" iterated="y" dur="2"/>

point-one-eight-seven about point-one-nine <pause dur="1.0"/> metres <pause dur="0.3"/> always when you get to this stage think about whether that seems reasonable or sort of point-one-nine metres twenty centimetres that kind of looks like a thickness <pause dur="0.3"/> of a wall <pause dur="0.4"/> if you know you're insulating a high temperature furnace and you get a wall thickness of a few millimetres then you suspect you've gone wrong <pause dur="0.5"/> # equally if you get sort of ten metres then <pause dur="0.3"/> you think this probably isn't realistic so you go back again and calculate but <pause dur="0.2"/> certainly that looks like <pause dur="0.3"/> a reasonable number <pause dur="1.1"/> so just doing the same for the other <pause dur="0.2"/> # <pause dur="0.2"/> # insulation material <pause dur="0.3"/> the fibreglass <pause dur="3.3"/><kinesic desc="writes on board" iterated="y" dur="15"/> exactly the same calculation <pause dur="0.2"/> but X-B <pause dur="1.8"/> is therefore equal to the temperature difference across the fibreglass four-hundred minus fifty-five <pause dur="1.3"/> degrees kelvin <pause dur="0.2"/> or celsius <pause dur="0.4"/> multiplied by the thermal conductivity of the fibreglass <pause dur="0.3"/> point-zero-four <pause dur="2.1"/><kinesic desc="writes on board" iterated="y" dur="12"/> watts per metre per degree kelvin <pause dur="8.7"/> so a better insulation material <pause dur="0.3"/> and again divided by two-hundred-and-twenty-five <pause dur="8.5"/><kinesic desc="writes on board" iterated="y" dur="8"/>

which is equal to </u><pause dur="17.0"/> <u who="sm0848" trans="pause"> zero-six <gap reason="inaudible" extent="1 sec"/></u><u who="nf0831" trans="overlap"> <trunc>part</trunc> point </u><pause dur="0.4"/> <u who="sm0848" trans="pause"> zero-six </u><u who="nf0831" trans="overlap"> that sounds about right <pause dur="0.7"/> i was going to say was it's something like <pause dur="0.2"/> one-and-a-half times <pause dur="0.4"/> point-zero-four so that's about point-zero-six <pause dur="3.1"/><kinesic desc="writes on board" iterated="y" dur="2"/> metres so again sort of numbers that look as if you could build <pause dur="0.3"/> an insulation system out of it </u><u who="sm0849" trans="latching"> what are they <gap reason="inaudible" extent="2 secs"/></u><pause dur="0.4"/> <u who="nf0831" trans="pause"> sorry </u><pause dur="0.2"/> <u who="sm0849" trans="pause"> what are they <gap reason="inaudible" extent="1 sec"/> <pause dur="0.5"/> blocks <pause dur="1.4"/> <gap reason="inaudible" extent="1 sec"/> wall blocks </u><u who="nf0831" trans="overlap"> wall blocks <pause dur="0.4"/> they're sort of <trunc>s</trunc> # compressed <pause dur="0.2"/> # i mean they're they're solid insulation <pause dur="0.5"/> blocks <pause dur="0.4"/> that you use for high temperature <pause dur="0.2"/> # sort of a first <pause dur="0.3"/> <trunc>li</trunc> lining layer you you <pause dur="0.3"/> you would actually have a steel layer <pause dur="0.8"/> or you you would have <pause dur="0.4"/> a steel layer or <unclear>somewhere</unclear> a firebrick layer on the very <pause dur="0.3"/> inside <pause dur="0.5"/> and then they're they're sort of lightweight fibre <pause dur="0.4"/> compressed fibre blocks <pause dur="0.8"/> that you can <pause dur="0.6"/> build as a sort of an insulation wall <pause dur="2.0"/>

okay <pause dur="0.2"/> so <pause dur="0.2"/> that's <pause dur="0.2"/> big scale high temperature conventional engineering <pause dur="0.3"/> # sort of calculation <pause dur="1.1"/> # <pause dur="0.2"/> and the key thing is to break it down into the you know look <trunc>a</trunc> look at the problem look at the bits where you've got the information <pause dur="0.4"/> decide <pause dur="0.3"/> where <pause dur="0.2"/> you need to sort of go first to find the information <pause dur="0.3"/> you don't have <pause dur="1.6"/> the next thing i want to do and we won't probably complete it in the we'll complete it in the lecture next week <pause dur="0.8"/> is to go back to this problem we've started to look at in the last lecture <pause dur="0.6"/> on designing whether you can design a <pause dur="0.2"/> little portable <pause dur="0.4"/> # chilling system <pause dur="0.4"/> # so that the department store can <pause dur="0.5"/> present flowers or strawberries <pause dur="0.4"/> in its <pause dur="0.2"/> # # appropriate place to encourage people to <pause dur="0.3"/> buy them <pause dur="0.5"/> so this is example five-six in your notes <pause dur="5.2"/><kinesic desc="writes on board" iterated="y" dur="3"/> and this is completely open-ended # i <trunc>ha</trunc> i deliberately hadn't <pause dur="0.3"/> done the calculations before we do this we may end up that

it's completely <pause dur="0.2"/> not a good idea you can't it won't work <pause dur="0.4"/> # and so <pause dur="0.3"/> # you go back and think about something else <pause dur="0.6"/> so what we decided <pause dur="0.5"/> is that this sort of <kinesic desc="writes on board" iterated="y" dur="6"/> chiller unit <pause dur="6.2"/> is going to be made it's going to be <pause dur="0.2"/> cylindrical so that <pause dur="0.2"/> people can sort of reach in easily <pause dur="8.9"/><kinesic desc="writes on board" iterated="y" dur="6"/> and it's going to have <pause dur="0.6"/><kinesic desc="writes on board" iterated="y" dur="5"/> insulation on the outside we've <trunc>in</trunc> initially trying <pause dur="1.5"/> something like the insulation you'd put on a refrigerator <pause dur="0.3"/> so about sixty <kinesic desc="writes on board" iterated="y" dur="16"/> millimetres <pause dur="14.7"/> we'd taken a total height <pause dur="0.2"/> in the lecture last week of one metre <pause dur="6.8"/><kinesic desc="writes on board" iterated="y" dur="4"/> # a total external <pause dur="0.2"/><kinesic desc="writes on board" iterated="y" dur="6"/> diameter <pause dur="0.5"/> of six-hundred millimetres <pause dur="5.3"/> and an insulation <pause dur="0.2"/><kinesic desc="writes on board" iterated="y" dur="17"/> thickness <pause dur="0.6"/> of <pause dur="0.2"/> sixty <pause dur="0.4"/> millimetres </u><pause dur="9.3"/> <u who="sm0850" trans="pause"> does that bit need to be sixty as well <pause dur="0.7"/> that last one you've just drawn </u><pause dur="0.3"/> <u who="nf0831" trans="pause"> sorry </u><pause dur="0.4"/> <u who="sm0850" trans="pause"> that last one you've just drawn </u><pause dur="0.6"/> <u who="nf0831" trans="pause"> this last line i've not drawn doesn't matter this last line i have drawn doesn't need to be sixty <pause dur="0.5"/><kinesic desc="writes on board" iterated="y" dur="15"/> because this thickness here there's ice in here <pause dur="2.6"/> and there's the fresh produce <pause dur="0.2"/> sitting on top <pause dur="1.3"/><event desc="changes pen" iterated="n"/> think i've got a a fresh produce colour that i can <pause dur="0.3"/> draw in <pause dur="9.9"/><kinesic desc="writes on board" iterated="y" dur="3"/> don't think i can draw anything like a rose so we'll have a sort of a <pause dur="0.9"/> # <pause dur="0.2"/> # <pause dur="0.2"/> nursery flower <pause dur="0.7"/> and the the condition here <pause dur="0.4"/>

is that you want you don't want it to be so cold here that the bottoms of the flowers are going to freeze <pause dur="0.4"/> so we've said that the condition <kinesic desc="indicates point on board" iterated="n"/> here <pause dur="0.8"/> is <pause dur="0.3"/> that <pause dur="0.2"/> the <pause dur="0.6"/> temperature <pause dur="0.6"/> at this point <pause dur="0.9"/> should be <pause dur="0.5"/> four <pause dur="0.2"/> degrees </u><pause dur="0.8"/> <u who="sm0851" trans="pause"> so we've got one work that out </u><pause dur="0.3"/> <u who="nf0831" trans="pause"> so </u><pause dur="0.8"/> <u who="sm0852" trans="pause"> that's too high </u><pause dur="0.4"/> <u who="nf0831" trans="pause"> that's <pause dur="1.6"/> the first <pause dur="0.2"/> calculation to do is to find the thickness <kinesic desc="indicates point on board" iterated="n"/><pause dur="0.3"/> of <pause dur="0.2"/> that bit <pause dur="1.3"/> the ice is in here <pause dur="0.8"/> and again first approximation we'll assume that it stays exactly at zero degrees as it melts <pause dur="0.5"/> which if you look at real melting ice is <pause dur="0.2"/> almost true <pause dur="0.3"/> so we've got ice in <kinesic desc="indicates point on board" iterated="n"/> here that we hope keeps this system cold <pause dur="0.3"/> throughout the working day <pause dur="2.0"/> we've got sixty millimetres of insulation round all the other <pause dur="0.2"/> surfaces <pause dur="0.4"/><kinesic desc="writes on board" iterated="y" dur="4"/> of the ice <pause dur="6.4"/> and <pause dur="0.3"/> we'd established that to <pause dur="0.5"/> put the produce in <pause dur="0.4"/> we need <kinesic desc="writes on board" iterated="y" dur="11"/> probably a height of about three-hundred millimetres <pause dur="0.5"/> here <pause dur="9.2"/> okay like all real problems you look at that and you say that's far too difficult to do i can't do that <pause dur="0.4"/> # and # <pause dur="0.2"/> so <pause dur="0.3"/> <trunc>y</trunc> you have to look at it and decide what you think are going to be the <pause dur="0.5"/> key <pause dur="0.3"/> # <pause dur="0.2"/> factors that determine the rate of

heat transfer <pause dur="0.3"/> so that you can do your first calculation <pause dur="0.5"/> 'cause your first calculation says that you're going to need ten tons of ice <pause dur="0.3"/> to achieve this <pause dur="0.3"/> then you know it's not going to work <pause dur="0.2"/> whatever you do <pause dur="0.3"/> in your calculations <pause dur="0.2"/> and so you go away and think about something else <pause dur="0.3"/> if your first calculation says yes this looks quite a good idea <pause dur="0.4"/> then you think about doing the calculation better <pause dur="0.4"/> to understand the physical system <pause dur="0.3"/> better <pause dur="1.3"/> so <pause dur="0.5"/> can you suggest any bit of this that we can ignore 'cause in principle we've got heat transfer here we've got some complicated heat transfer <pause dur="0.4"/> through these sides <pause dur="0.3"/> got heat transfer from the ice across the <pause dur="0.5"/> probably thin layer of insulation here <pause dur="0.3"/> heat transfer from the ice <pause dur="0.5"/> across to the air round here <pause dur="0.7"/> and also <pause dur="0.2"/> this is probably sitting on the <pause dur="1.1"/> nice plush carpet in the <pause dur="0.5"/> soft <pause dur="0.2"/> furnishings part of the <pause dur="0.2"/> store <pause dur="0.3"/> so we've got some heat transfer by conduction <pause dur="0.4"/> down <pause dur="0.2"/> to the <pause dur="0.3"/> ground or sorry conduction actually <pause dur="0.2"/> in positive heat transfer

the heat is all coming in to the ice which is gradually <pause dur="0.2"/> melting </u><u who="sm0852" trans="latching"> can you not ignore that if it's on wheels </u><pause dur="0.2"/> <u who="nf0831" trans="pause"> sorry </u><pause dur="0.2"/> <u who="sm0852" trans="pause"> can you not ignore that if it's on wheels </u><pause dur="0.5"/> <u who="nf0831" trans="pause"> that's a good point so if we if we put it on wheels <pause dur="0.4"/> in fact it makes <trunc>sligh</trunc> a calculation slightly more <pause dur="0.2"/> more you're you're # you're putting in an extra calculation so we'll put it on wheels <pause dur="2.9"/><kinesic desc="writes on board" iterated="y" dur="2"/></u><pause dur="0.6"/> <u who="sm0853" trans="pause"> <gap reason="inaudible" extent="2 secs"/></u><pause dur="0.4"/> <u who="nf0831" trans="pause"> ah but do you need the sort of wheels that you're going to be able to sort of sit it down </u><u who="sm0853" trans="overlap"> no </u><u who="nf0831" trans="overlap"> so that people don't your customers don't start pushing it around so <pause dur="0.3"/> <vocal desc="laughter" n="sm0853" iterated="y" dur="2"/> perhaps you need some sort of retractable <kinesic desc="writes on board" iterated="y" dur="2"/> castors that <pause dur="0.7"/> will sit in here so that your customers don't wheel the strawberries out of the store with their <pause dur="0.3"/><vocal desc="laughter" n="ss" iterated="y" dur="2"/> trollies <pause dur="1.4"/> so we'll we'll put it back on the carpet i think <pause dur="1.2"/> that makes it easier 'cause if you've got a thick layer of insulation <pause dur="0.4"/> and a thick layer of carpet <pause dur="0.3"/> and some flooring here <pause dur="0.4"/> my first sort of approximation was going to say let's forget about conduction to the floor <pause dur="0.3"/> to begin with <pause dur="0.3"/> because <pause dur="0.3"/> probably <pause dur="0.4"/> the <pause dur="0.2"/> transfer <pause dur="0.3"/> from the surface is going to be much more rapid than the <pause dur="0.2"/> transfer

through sort of whatever is down here <pause dur="0.3"/> and that could be the next stage that you go back and say well was that important <pause dur="0.3"/> or <pause dur="0.3"/> not <pause dur="0.5"/> so my sort of first approach on this would <trunc>a</trunc> actually be to <pause dur="0.3"/> # quickly get the trolley off its wheels <pause dur="0.2"/> sitting on a a nice thick carpet and say <pause dur="0.3"/> # # sort of our first approximations <pause dur="4.9"/><kinesic desc="writes on board" iterated="y" dur="10"/> is to ignore the base <pause dur="5.6"/> anything else you think is going to be too difficult to worry about to begin with and probably isn't going to be <pause dur="0.3"/> # very important </u><pause dur="5.0"/> <u who="sm0854" trans="pause"> the sides above our ice </u><pause dur="0.4"/> <u who="nf0831" trans="pause"> yes i thought those will have <trunc>tho</trunc> those started to look very # <pause dur="0.3"/> difficult because there's going to be a sort of a temperature <kinesic desc="indicates point on board" iterated="n"/> gradient up here and we're not quite sure <pause dur="0.4"/> # at all what's going to be in there <pause dur="0.4"/> so i think another very good approximation is to forget about these sides as a first approximation you might want to come back <pause dur="0.3"/> and do it later <pause dur="0.9"/><kinesic desc="writes on board" iterated="y" dur="3"/> so if we now ignore the sides above <pause dur="0.2"/> the level <pause dur="0.3"/> of the insulation <pause dur="9.3"/><kinesic desc="writes on board" iterated="y" dur="8"/> # <pause dur="0.2"/> i need a word to describe this if <trunc>w</trunc> if we <pause dur="0.4"/> call this the <pause dur="0.3"/> a <kinesic desc="writes on board" iterated="y" dur="7"/> shelf <pause dur="0.3"/> if you like <pause dur="7.1"/> then in

fact you end up with <pause dur="0.3"/> a much simpler problem <pause dur="0.5"/> to deal with <pause dur="0.7"/><kinesic desc="writes on board" iterated="y" dur="30"/> you've now got something that looks like this <pause dur="2.9"/> this is the shelf with the thickness that <pause dur="0.2"/> keeps the <pause dur="0.3"/> produce so it doesn't get too cold <pause dur="10.4"/> and this in cross-section <pause dur="0.3"/> is a cross-section through a much shorter <pause dur="0.5"/> cylinder <pause dur="2.5"/> and this is the ice <pause dur="1.8"/> at <pause dur="0.4"/> zero degrees <pause dur="0.6"/> celsius here <pause dur="0.9"/> so you've made the problem <pause dur="0.6"/> something <pause dur="0.3"/> that you've only now got <pause dur="0.2"/> two different surfaces to worry about <pause dur="0.5"/> there's the shelf <pause dur="5.6"/><kinesic desc="writes on board" iterated="y" dur="4"/> and then there's the cylindrical <pause dur="0.5"/> sides the short length of them <pause dur="10.1"/><kinesic desc="writes on board" iterated="y" dur="7"/> so that looks like the sort of thing that you've got the <pause dur="0.2"/> # tools to <pause dur="0.4"/> work with <pause dur="0.6"/> # and in fact the shelf bit is very much like the problem <pause dur="0.3"/> that we've looked just looked at <pause dur="0.5"/> so just getting down the basic <pause dur="0.2"/> information that you've got on this shelf <pause dur="0.5"/> you know it's six-hundred millimetres <pause dur="0.3"/> in external diameter and a wall thickness of sixty <pause dur="0.6"/> so that makes it <pause dur="0.2"/> # four-hundred-and-<pause dur="0.2"/>eighty thank you <pause dur="0.6"/><kinesic desc="writes on board" iterated="y" dur="13"/> millimetres <pause dur="0.7"/> diameter here <pause dur="5.0"/> we don't know the thickness <pause dur="0.4"/> that's the <pause dur="0.6"/> next thing to <pause dur="1.0"/>

calculate so that's the first bit of the calculation <pause dur="0.2"/> that we need to do next week <pause dur="2.4"/> because we don't know the thickness there we don't know <pause dur="0.2"/> the length down here <pause dur="0.7"/> although you would <pause dur="0.6"/> expect that that thickness is going to be quite small so we're going to be looking probably at something like seven-hundred <pause dur="0.4"/> # sorry <pause dur="0.4"/> seven-hundred millimetres less the <pause dur="0.4"/> because of the insulation there it's not going to be <pause dur="0.3"/> very large so we've got a sort of feel for <pause dur="0.3"/> what that height will be <pause dur="1.2"/> and so what we'll do in the next lecture <pause dur="0.8"/> is to think about how we can describe convection and radiation from the surface

here <pause dur="0.6"/> # <pause dur="0.3"/> initially thinking about this just as being a sort of a bare plate which is the worst case <pause dur="0.6"/> # <pause dur="0.2"/> and then look at the <pause dur="0.2"/> convection and radiation <pause dur="0.3"/> and conduction through <pause dur="0.2"/> the <pause dur="0.2"/> sides <pause dur="0.5"/> i'll then finish off next week with the final calculation <pause dur="0.3"/> on <pause dur="0.2"/> the # # a sort of a a complicated system of a double glazed window <pause dur="1.1"/> okay <pause dur="0.7"/> Friday is test day <pause dur="0.4"/> please don't forget that <pause dur="0.8"/> # <pause dur="0.2"/> so come with writing implements <pause dur="0.3"/> # pen not red pen please <pause dur="0.3"/> # <pause dur="0.2"/> diagrams in pencil but writing in pen <pause dur="0.5"/> and also your university approved calculator <pause dur="0.6"/> okay thank you